Assume that U{\displaystyle U} is a bounded subset of Rn{\displaystyle \mathbb {R} ^{n}} with smooth boundary and use the weak formulation of the problem to prove the existence of a unique weak solution.

Consider the functional B[u,v]=∫UΔuΔv=∫Ufv{\displaystyle B[u,v]=\int _{U}\Delta u\Delta v=\int _{U}fv}. B{\displaystyle B} is bilinear by linearity of the Laplacian. Now, we claim that B{\displaystyle B} is also continuous and coercive.

|B[u,v]|=|∫UΔuΔv|≤∥Δu∥L2(U)∥Δv∥L2(U)≤∥Δu∥H01(U)∥Δv∥H01(U){\displaystyle |B[u,v]|=\left|\int _{U}\Delta u\Delta v\right|\leq \|\Delta u\|_{L^{2}(U)}\|\Delta v\|_{L^{2}(U)}\leq \|\Delta u\|_{H_{0}^{1}(U)}\|\Delta v\|_{H_{0}^{1}(U)}} where the first inequality is due to Holder and the second is by the definition of the Sobolev norm. And so B[u,v]{\displaystyle B[u,v]} is a continuous functional.

To show coercivity, we use the fact that by two uses of integration by parts, ∫Uuijuij=−∫Uuiuijj=∫Uuiiujj{\displaystyle \int _{U}u_{ij}u_{ij}=-\int _{U}u_{i}u_{ijj}=\int _{U}u_{ii}u_{jj}} which gives